Since every 10 X 10 Latin square has an embedded 3 X 3 Latin square, could we seed three rows, three columns, and their intersections as the 3 X 3 Latin square?
Could your algorithm be modified to start with this additional information?

Since every 10 X 10 Latin square has an embedded 3 X 3 Latin square, could we seed three rows, three columns, and their intersections as the 3 X 3 Latin square?
Could your algorithm be modified to start with this additional information?

Could you elaborate more about this? It is unclear where exactly these 3x3 squares should be placed.

I thought about possibility to reuse existing squares and found another promising approach. You should start with existing ODLS pair of rank 8. Take first square from pair and extend it to rank 10 by appending rows and columns around this square. Then permute rows 2-9 of new square in the same way as in 2nd square from pair. You can also swap first and last rows. This looks like a promising way to find rank 10 ODLS pair.

Since every 10 X 10 Latin square has an embedded 3 X 3 Latin square, could we seed three rows, three columns, and their intersections as the 3 X 3 Latin square?
Could your algorithm be modified to start with this additional information?

Now, used the next scheme of cells filling:
If I understand right, you ask about the next scheme:
Algorithm of search can be adjusted for a scheme like showed on the second image. An idea like "get a combination of ODLS of rank n and build a combination (pair, triple, e.t.c.) of rank n+1", as I know, tested previously (not in RakeSearch) but the result was disappointing. A simple example: we know pairs of orthogonal Latin squares of rank 5. But does not exists any pair of Latin squares of rank 6. But many other schemes of building orthogonal pairs of Latin squares (diagonal and usual) exists and tested for example, in Gerasim@Home project.

Copy of this message by request of Michael:
As tell previously, we add some explanations about the new run of R10 search. We expect that "permutational pairs" of rank 10 - are very rare or do not even exist. In this case, 99.999% (or even 100%) results from computers contain information about a number of processed squares only. Not interesting and can hide potential errors in the algorithm. In a new application for the current run, we add to results files not the only number of processed squares and orthogonal pairs (if they are found), but partial orthogonal pairs also.
Sample of similar pair from the result, processed on my computer:
Top left square - a diagonal Latin square created during sequential generation. Top right square - generated from the first square by rows permutation. Bottom square with a two-digit number in each cell - Graeco-Latin square - the combination of two bottom squares, which were superimposed on each other. If top squares are orthogonal, all 100 two-digit numbers from Graeco-Latin square was distinct. But in this case, it is not. Only 64 values inside cells - unique (they marks by black), and other 17 numbers - present in this square more than once. If we count of distinct numbers in the bottom square we will get 81 - this is the degree of orthogonality of top pair of squares!
In the new run, we place in results all pairs with a degree of orthogonality > 80. And now, in each confirmed result we see a several or even tens partially orthogonal pairs - sometimes more than 30, for example. May be each result contain interesting information. Not ODLS, but interesting also.

2n X 2n Latin squares are problematic when n is odd, i.e., 6 X 6 does not exist, 10 X 10 was big news when it was found to contain a 3 X 3 --- it made the cover of Scientific American.
Unfortunately, I don't have the knowledge to contribute more. Good luck.

2n X 2n Latin squares are problematic when n is odd, i.e., 6 X 6 does not exist, 10 X 10 was big news when it was found to contain a 3 X 3 --- it made the cover of Scientific American.
Unfortunately, I don't have the knowledge to contribute more. Good luck.

You show it in message 1118, second panel.
It looks to be asymmetric, which might be the problem with 2n X 2n squares when n is odd.
Maybe you could move to 11 X 11, if the search space is reasonable. And do 10 X 10 as Beta.